Think of the three charges as laying on the x-axis with the left charge q1 at the origin. The field at any point along the x-axis is the vector sum of the individual fields that each charge would produce by itself. This is called the principle of
superposition. The fact that all charges lie along a straight line helps to simplify the situation.
(a). To calculate the electric field E produced by a set of i charges at some point P in space consider the following equation:
http://home.comcast.net/~perion_666/Files/Eq-1.gif
E is calculated for a point in space given by the point's position vector R_p. For your problem we are only dealing with some point P located "1 cm (0.01 meters) to the left of the middle charge".
You have to watch your units! Try to express evrything in SI units (Meters, Coulombs, Newtons, etc). So you'll have to express all your distances as meters rather than cm and charge in Coulombs rather than uC. Epsilon sub 0 is the "permittivity of free space" constant which in SI units is about 8.85 x 10^-12 C^2/Nm^2
r_i is the distance (in meters) between each charge q_i and P. Vector u_i is the unit vector in the direction
from the charge
to P. Since each charge is positioned on the x axis, each unit vector will equal either plus or minus 1, depending on whether P is to left (minus) or right (plus) of the particular charge being evaluated.
So, if q1 is located at x=0, then your point P is at x=0.02 meters, q2 is at x=0.03 meters, and q3 is at x=0.05 meters. From these you can figure the distances r_i between each q and P and the correct signs for each of the unit vectors
from each q
to P. Furthermore, q1 and q2 are positive charges while q3 is negative so use those signs for each q in its calculation. Each E field strength vector is calculated for each charge of three charges and they are all added together: E(r) = E(1) + E(2) + E(3) where E(1) is the calculation using q1, etc.
(b). Now that you've calculated the field strength vector at P, find the force it would have on the -3.50 uC (-3.5 x 10^-6 Coulombs) test charge by multiplying the value of E at P by the value of the charge: i.e.
F = qE . The resulting force is in Newtons if you kept all your units constant.
Remember, the Electric field is a value that gives the
force per unit charge at a point in space without reference to any test charge actually existing at that point. If a charge is actually inserted there it's a matter of multiplying that charge value by the force per unit charge to arrive at the actual force exerted on the test charge.
Apologies if this is confusing and I hope I haven't stated something erroneously - someone will correct me, hopefully
